Exercise E6.3 Lines and Angles NCERT Solutions Class 9

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Question 1

In Fig. below, sides \(QP\) and \(RQ\) of \(∆ PQR\) are produced to points \(S\) and \(T\) respectively. If \(\angle SPR = 135^\circ \) and \(\angle PQT = 110^\circ \) , find \(\angle PRQ\).

Solution

What is known?

\(\angle SPR = 135^\circ \) and \(\angle PQT = 110^\circ \)

What is unknown?

\(\angle PRQ\)

Reasoning:

As we know the linear pair axioms:

If Non-common arms of two adjacent angles form a line, then these angles are called linear pair of angles and the sum of the linear pair is \(180^\circ\).

If the sum of two adjacent angles is \(180^\circ\) then the two non-common arms of the angles form a line.

Angle sum property of a triangle:

Sum of the interior angles of a triangle is \(360^\circ\).

Steps:

Given,

\(\angle SPR = 135^\circ \) and \(\angle PQT = 110^\circ \)

\(\begin{align}\angle SPR + \angle QPR &= 180^\circ \\ {\text{By linear pair axiom}} \\\\135^\circ + \angle QPR &= 180^\circ \\\angle QPR &=180^\circ\! -135^\circ \\\angle QPR &= 45^\circ.... {\rm{ }}\left( {\rm{i}} \right)\end{align}\)

\(\begin{align}\angle PQT + \angle PQR &= 180^\circ \\\begin{array}{I}(\text{By linear pair axiom}) \end{array}\\\\110^\circ + \angle PQR &= 180^\circ \\\angle PQR &= 180^\circ - 110^\circ \\\angle PQR &= 70^\circ..\left( {{\rm{ii}}} \right)\end{align}\)

Now,

\(\begin{align}\angle PQR + \angle QPR + \angle PRQ &= 180^\circ\\ (\text{Angle sum } \text{property of a}&\text{ triangle}) \\\\70^\circ + {\rm{ }}45^\circ + \angle PRQ &= 180^\circ\\{\rm{}}\left[ {{\rm{from }}\left( {\rm{i}} \right){\rm{ and }}\left( {{\rm{ii}}} \right)} \right]\\\\\angle PRQ &\!=180^\circ\!\! -\!115^\circ \\\angle PRQ& = 65^\circ \end{align}\)

Question 2

In Fig. below, \(\angle X = 62^\circ ,{\rm{ }}\angle XYZ = 54^\circ \). If \(YO\) and \(ZO\) are the bisectors of \(\angle XYZ\) and \(\angle XZY\) respectively of \(\Delta XYZ,\) find \(\angle OZY\) and \(\angle YOZ\) .

Solution

What is known?

\(\angle X = 62^\circ ,{\rm{ }}\angle XYZ = 54^\circ \) and \(YO\) and \(ZO\) are the bisectors of \(\angle XYZ\) and \(\angle XZY\) respectively.

What is unknown?

\(\angle OZY\) and \(\angle YOZ\)

Reasoning:

As we know the angle sum property of a triangle:

Sum of the interior angles of a triangle is \(360^\circ\).

Steps:

Given, in \(\Delta XYZ\),

\[\begin{align}\angle X &= 62^\circ \\\angle XYZ &= 54^\circ \end{align}\]

\(\begin{align}\angle X + \angle XYZ + \angle Z &= 180^\circ\\ ( \text{Angle sum property of }&\text{a triangle}\rm{.} )\\\\62^\circ + {\rm{ 5}}4^\circ + \angle Z &= 180^\circ \\\angle Z &=180^\circ-116^\circ \\\angle Z &= 64^\circ \end{align}\)

Now, \(OZ\) is angle bisector of \(\angle XZY\)

\(\begin{align} \angle OZY = \frac{1}{2}\angle XZY &= \frac{1}{2} \times 64^\circ \\&= 32^\circ {\rm{   }}\left( {\rm{i}} \right)\end{align} \)

Similarly, \(OY\) is angle bisector of \(\angle XYZ\)

\(\begin{align} \angle OYZ = \frac{1}{2}\angle XYZ &= \frac{1}{2} \times 54^\circ \\ &= 27^\circ \left( {{\rm{ii}}} \right)\end{align} \)

Now, in \(\Delta OYZ\)

\(\begin{align}\angle OYZ \!+\! \angle OZY \!+\! \angle YOZ &= 180^\circ \\( \text{Angle sum property of a }&\text{triangle}\rm{.} )\\\\27^\circ +32^\circ + \angle YOZ & = 180^\circ \\\left[ {{\textrm{from }}\left( {\rm{i}} \right){\rm{ and }}\left( {{\rm{ii}}} \right)} \right]\\\\\angle YOZ &\!= 180^\circ \!- 59^\circ \\\angle YOZ &= 121^\circ \end{align}\)

Hence, \(\angle OZY = 32^\circ \) and \(\angle YOZ = 121^\circ \)

Question 3

In Fig. below, if \(AB || DE\), \(\angle BAC = 35^\circ \) and \(\angle CDE = 53^\circ \), find \(\angle DCE\).

Solution

What is known?

\(AB || DE\), \(\angle BAC = 35^\circ \) and \(\angle CDE = 53^\circ \)

What is unknown?

\(\angle DCE\)

Reasoning:

As we know when two parallel lines are cut by a transversal, alternate interior angles formed are equal.

Angle sum property of a triangle:

Sum of the interior angles of a triangle is \(360^\circ\).

Steps:

Given,

\(AB || DE\) , \(\angle BAC = 35^\circ \) and \(\angle CDE = 53^\circ \)

\(\begin{align}\angle DEC &= \!\angle BAC\\( \text{Alternate }&\text{interior angles})\\\\\angle DEC &= 35^\circ \end{align}\)

Now, in \(\Delta CDE\)

\(\begin{align}\angle CDE + \angle DEC + \angle DCE &= 180^\circ\quad \\( \text{Angle sum property of a}&\text{ triangle.})\\\\53^\circ + 35^\circ + \angle DCE &= 180^\circ \\\angle DCE &\!= 180^\circ \! - 88^\circ \\\angle DCE &= 92^\circ\end{align}\)

Question 4

In fig. below, if lines \(PQ\) and \(RS\) intersect at point \(T\), such that \(\angle PRT = 40^\circ \) ,\(\angle RPT = 95^\circ \) and \(\angle TSQ = 75^\circ \) , find \(\angle SQT\) .

Solution

What is known?

\(\angle PRT = 40^\circ \), \(\angle RPT = 95^\circ \) and \(\angle TSQ = 75^\circ \)

What is unknown?

\(\angle SQT\)

Reasoning:

As we know when two line intersect each other at a point then there are two pairs of vertically opposite angles formed are equal.

Angle sum property of a triangle:

Sum of the interior angles of a triangle is \(360^\circ\).

Steps:

Given,

\(\angle PRT = 40^\circ \), \(\angle RPT = 95^\circ \) and \(\angle TSQ = 75^\circ \)

In \(\Delta PRT\)

\(\begin{align}\angle PTR + \angle PRT + \angle RPT &= 180^\circ {\rm{ }}\\( \text{Angle sum property of a}&\text { triangle}\rm{.})\\\\\angle PTR + 40^\circ + 95^\circ &= 180^\circ \\\angle PTR &= 180^\circ - 135^\circ\\\angle{PTR} &= 45^\circ {\rm{ }}\end{align}\)

Now,

\(\begin{align}\angle QTS& = \angle PTR\quad\\( \text{Vertically }&\text {opposite angles})\\\\\angle QTS &= 45^\circ\quad \left( {\rm{i}} \right)\end{align}\)

In \(\Delta TSQ\)

\(\begin{align}\angle QTS + \angle TSQ + \angle SQT &= 180^\circ\\( \text{Angle sum property of a }&\text {triangle}.)\\\\45^\circ + 75^\circ + \angle SQT &= 180^\circ \\ [ \text{From }&( \rm{i} ) ]\\\\\angle SQT &= 180^\circ - 120^\circ \\\angle SQT &= 60^\circ \end{align}\)

Hence,\(\angle SQT = 60^\circ {\rm{ }}\)

Question 5

In Fig. below, if \(PQ ⊥ PS\) , \(PQ || SR\) ,\(\angle SQR = 28^\circ \) and \(\angle QRT = 65^\circ \) ,then find the values of \(x\) and \(y\) .

Solution

What is known?

\(PQ ⊥ PS,\, PQ || SR\),  \(\angle SQR = 28^\circ \) and \(\angle QRT = 65^\circ \)

What is unknown?

\(x\) and \(y\).

Reasoning:

As we know when two parallel lines are cut by a transversal, alternate interior angles formed are equal.

Angle sum property of a triangle:

Sum of the interior angles of a triangle is \(360^\circ\).

Steps:

Given,

\(PQ ⊥ PS,\, PQ || SR\),  \(\angle SQR = 28^\circ \) and \(\angle QRT = 65^\circ \)

\(\begin{align}\angle PQR &= \angle QRT\\ \begin{array}{I}(\text{Alternate }\end{array} &\begin{array}{I}\text{interior angles}) \end{array}\\\\ \angle PQS + \angle SQR &= \angle QRT \\ \text{(By figure)}\\\\x + 28^\circ &= 65^\circ \\x &= 65^\circ - 28^\circ \\x &= 37^\circ \end{align}\)

Now, in \(\Delta PQS\)

\(\begin{align}\angle PQS + \angle PSQ + \angle QPS &=180^\circ\\ ( \text{Angle sum property of a }& \text{triangle}\rm{.})\\ \\37^\circ + y + 90^\circ &= 180^\circ \\y &=180^\circ \!- \!127^\circ \\y &=\, 53^\circ \end{align}\)

Hence, \(x = 37^\circ \) and \(y = 53^\circ \)

Question 6

In Fig. below, the side \(QR\) of \(\Delta PQR\) is produced to a point \(S\). If the bisectors of \(\angle PQR\) and \(\angle PRS\) meet at point \(T\), then prove that \(\begin{align} \angle QTR = \frac{1}{2}\angle QPR\end{align} \) .

Solution

What is known?

The side \(QR\) of \(\Delta PQR\) is produced to a point \(S\) and the bisectors of \(\angle PQR\) and \(\angle PRS\) meet at point \(T\).

To prove:

\(\begin{align}\angle QTR = \frac{1}{2}\angle QPR\end{align}\)

Reasoning:

As we know exterior angle of a triangle:

If a side of a triangle is produced, then the exterior angle so formed is equal to the sum of the two interior opposite angles.

Steps:

Given,

Bisectors of \(\angle PQR\) and \(\angle PRS\) meet at point \(T\).

Hence, \(TR\) is a bisector of \(\angle PRS\) and \(TQ\) is a bisector of \(\angle PQR\) 

\(\begin{align} \angle PRS &= 2\angle TRS\quad\dots\left( {\rm{i}} \right)\\\angle PQR &= 2\angle TQR\quad\dots\left( {{\rm{ii}}} \right)\end{align} \)

Now, in \(\Delta TQR\)

\(\begin{align}\angle TRS&= \angle TQR + \angle QTR\\ &\left( {{\text{Exterior angle of triangle}}} \right)\\\\\angle QTR &= \angle TRS-\angle TQR\quad\dots\left( {{\text{iii}}} \right)\end{align}\)

Similarly, in \(\Delta PQR\)

\(\begin{align}\angle PRS &= \angle QPR + \angle PQR\qquad\\ &\left( {{\text{Exterior angle of triangle}}} \right)\\\\2\angle TRS &= \angle QPR + 2\angle TQR\\ &\left[ {{\text{From }}\left( {\rm{i}} \right){\rm{ and }}\left( {{\rm{ii}}} \right)} \right]\\\\\angle QPR &= 2\angle TRS-2\angle TQR\\\angle QPR &= 2\left( {\angle TRS-\angle TQR} \right)\\\angle QPR &= 2\angle QTR\qquad\qquad\,\,\,\,\,\qquad\\ &\left[ {{\text{From }}\left( {{\rm{iii}}} \right)} \right]\\\\\angle QTR &= \frac{1}{2}\angle QPR\end{align}\)

Hence proved.

  
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